a)The FBD indicates the ball-and-socket support provides _____ reaction(s). b)The position vector from the ball-and-socket support to the load has a magnitude of _____ m. c)The length of the unit vector AB has a magnitude of _____ m.
a)The FBD indicates the ball-and-socket support provides _____ reaction(s). b)The position vector from the ball-and-socket support to the load has a magnitude of _____ m. c)The length of the unit vector AB has a magnitude of _____ m.
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
Related questions
Question
a)The FBD indicates the ball-and-socket support provides _____ reaction(s).
b)The position vector from the ball-and-socket support to the load has a magnitude of _____ m.
c)The length of the unit vector AB has a magnitude of _____ m.
![SAMPLE PROBLEM 3/7
The welded tubular frame is secured to the horizontal r-y plane by a ball-
and-socket joint at A and receives support from the loose-fitting ring at B. Under
the action of the 2-kN load, rotation about a line from A to B is prevented by the
cable CD, and the frame is stable in the position shown. Neglect the weight of
the frame compared with the applied load and determine the tension T in the
cable, the reaction at the ring, and the reaction components at A.
2.5
4,5m
2 kN
6 m
25 m
Solution. The system is clearly three-dimensional with no lines or planes of
symmetry, and therefore the problem must be analyzed as a general space sys-
tem of forces. The free-body diagram is drawn, where the ring reaction is shown
in terms of its two components. All unknowns except T may be eliminated by a
moment sum about the line AB. The direction of AB is specified by the unit
1m
(4.5j + 6k) - (3j + 4k). The moment of T about AB
vector n-
B.
6 + 4.5
is the component in the direction of AB of the vector moment about the point A
and equals r, x T.n. Similarly the moment of the applied load F about AB is
r, x F-n. With CD - 46.2 m, the vector expressions for T, F, r, and r, are
T-.
(21 + 2.5 - 6k)
46.2
F- 23 kN
r, - -i+ 2.5j m
r- 2.51 + 6k m
The moment equation now becomes
B
[EMA - 01 (-1+ 2.5) -
T
(21 + 2.5) - 6k) 3 + 4k)
46.2
+ (2.51 + 6k) x (2) (3j + 4k) - 0
Completion of the vector operations gives
48T
+ 20 - 0
46.2
T- 2.83 kN
Ans.
Helpful Hints
and the components of T become
The advantage of using vector nota-
tion in this problem is the freedom
to take moments directly about any
axis. In this problem this freedom
permita the choice of an axis that
eliminates five of the unknowns.
T, - 0.833 kN
T, - 1.042 kN
T, - -2.50 kN
We may find the remaining unknowns by moment and force summations as
follows:
(EM, - 0]
2(2.5) - 4.58, - 1.042(3) - 0
B, - 0.417 kN
Ans.
Recall that the vector r in the expres-
sion r x F for the moment of a force
is a vector from the moment center to
(EM, - 0]
4.5B, - 2(6) - 1.042(6) - 0
B, - 4.06 kN
Ans.
any point on the line of action of the
force. Instead of r, an equally simple
choice would be the vector AC.
(LF, - 01
A, + 0.417 + 0.833-0
A, --1.250 kN
Ans.
(LF, - 0)
A, + 2 + 1.042 -0
A, - -3.04 kN
Ans.
(EF, - 0]
The negative signs associated with
the A-components indicate that they
are in the opposite direction to those
shown on the free-body diagram.
A, + 4.06 - 2.50 - 0
A,- -1.556 kN
Ans.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fae5c3a0e-bbda-462f-b7bc-315af0d1db18%2F84d1e20e-e61e-4fa1-985a-63bda37c535c%2Fx1k3f_processed.png&w=3840&q=75)
Transcribed Image Text:SAMPLE PROBLEM 3/7
The welded tubular frame is secured to the horizontal r-y plane by a ball-
and-socket joint at A and receives support from the loose-fitting ring at B. Under
the action of the 2-kN load, rotation about a line from A to B is prevented by the
cable CD, and the frame is stable in the position shown. Neglect the weight of
the frame compared with the applied load and determine the tension T in the
cable, the reaction at the ring, and the reaction components at A.
2.5
4,5m
2 kN
6 m
25 m
Solution. The system is clearly three-dimensional with no lines or planes of
symmetry, and therefore the problem must be analyzed as a general space sys-
tem of forces. The free-body diagram is drawn, where the ring reaction is shown
in terms of its two components. All unknowns except T may be eliminated by a
moment sum about the line AB. The direction of AB is specified by the unit
1m
(4.5j + 6k) - (3j + 4k). The moment of T about AB
vector n-
B.
6 + 4.5
is the component in the direction of AB of the vector moment about the point A
and equals r, x T.n. Similarly the moment of the applied load F about AB is
r, x F-n. With CD - 46.2 m, the vector expressions for T, F, r, and r, are
T-.
(21 + 2.5 - 6k)
46.2
F- 23 kN
r, - -i+ 2.5j m
r- 2.51 + 6k m
The moment equation now becomes
B
[EMA - 01 (-1+ 2.5) -
T
(21 + 2.5) - 6k) 3 + 4k)
46.2
+ (2.51 + 6k) x (2) (3j + 4k) - 0
Completion of the vector operations gives
48T
+ 20 - 0
46.2
T- 2.83 kN
Ans.
Helpful Hints
and the components of T become
The advantage of using vector nota-
tion in this problem is the freedom
to take moments directly about any
axis. In this problem this freedom
permita the choice of an axis that
eliminates five of the unknowns.
T, - 0.833 kN
T, - 1.042 kN
T, - -2.50 kN
We may find the remaining unknowns by moment and force summations as
follows:
(EM, - 0]
2(2.5) - 4.58, - 1.042(3) - 0
B, - 0.417 kN
Ans.
Recall that the vector r in the expres-
sion r x F for the moment of a force
is a vector from the moment center to
(EM, - 0]
4.5B, - 2(6) - 1.042(6) - 0
B, - 4.06 kN
Ans.
any point on the line of action of the
force. Instead of r, an equally simple
choice would be the vector AC.
(LF, - 01
A, + 0.417 + 0.833-0
A, --1.250 kN
Ans.
(LF, - 0)
A, + 2 + 1.042 -0
A, - -3.04 kN
Ans.
(EF, - 0]
The negative signs associated with
the A-components indicate that they
are in the opposite direction to those
shown on the free-body diagram.
A, + 4.06 - 2.50 - 0
A,- -1.556 kN
Ans.
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